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Tardigrade
Question
Physics
If temperature of an object is 140 ° F , then its temperature in centigrade is
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Q. If temperature of an object is $140 \,{}^\circ F$ , then its temperature in centigrade is
NTA Abhyas
NTA Abhyas 2020
A
$105^\circ C$
0%
B
$32^\circ C$
0%
C
$140^\circ C$
25%
D
$60^\circ C$
75%
Solution:
Relation between $^\circ C$ and $^\circ F$ ,
$\frac{C}{5}=\frac{F - 32}{9}$
$\Rightarrow \, \frac{C}{5}=\frac{140 - 32}{9}$
$\Rightarrow C=60^\circ C$